1
An outcome is a complete result
For two dice the outcomes are ordered pairs, not sums. Choosing the wrong granularity is the first and most damaging error.
Lesson · Discrete Probability · The model
Sample Spaces and Events
A probability model is a set of outcomes and an assignment of probabilities to them. Almost every mistake in this course is a mistake about what the outcomes are.
S the SAMPLE SPACE: all possible outcomes
E ⊆ S the EVENT: the outcomes we are asking about
Outcomes must be MUTUALLY EXCLUSIVE and EXHAUSTIVE.
"Equally likely" is an assumption about S, not a fact.
Two objects, one assumption
Core ideas
What this lesson establishes
Getting the sample space right is most of the work. The arithmetic that follows is Combinatorics and Counting.
2
An event is a subset
"The sum is 7" is the set of six pairs that sum to 7. Set operations from Sets, Relations, and Functions apply unchanged: ∪ is "or", ∩ is "and", complement is "not".
3
Equally likely is a modelling choice
It is justified by symmetry of the physical situation, and it must be checked. Sums of two dice are not equally likely; pairs are.
The material
Setting up a model
Outcomes, events, and the choice of granularity.
Definition · Sample space, outcome, event
The sample space S is the set of all possible outcomes of an experiment. An outcome is one element of S; an event is a subset of S.
The outcomes must be mutually exclusive (no two happen together) and exhaustive (at least one happens), so that exactly one happens. Those are the two obligations a proof by cases carries in “Cases, Counterexamples, and Writing a Proof” in Logic and Proof — overlap nowhere, cover everything — and they fail in the same ways.
Example · Two dice, two models
Model A: S is the 36 ordered pairs (1,1) through (6,6), all equally likely. Model B: S is the 11 sums 2, …, 12. Both are legitimate sample spaces; only in model A are the outcomes equally likely. A sum of 7 arises six ways and a sum of 2 arises one way.
That is the granularity trap, and it produced a famous error: Cardano and later gamblers computed dice probabilities from the sums and got them wrong. The fix is to keep outcomes fine enough that symmetry justifies equal likelihood.
Events as sets
in words as a set
A or B A ∪ B
A and B A ∩ B
not A Ā = S \ A
A but not B A \ B
exactly one of A, B A △ B
A and B cannot both A ∩ B = ∅ (mutually exclusive)
Every identity from “Set Identities and Proofs” in Sets, Relations, and Functions therefore applies, and De Morgan is used constantly: "not (A or B)" is "neither A nor B", which is usually the easier event to count.
Definition · Probability of an event (equally likely case)
When the outcomes of a finite S are equally likely, P(E) = |E| / |S|. Every probability in this course's early lessons is this ratio, computed with Combinatorics and Counting's tools.
Example · Choosing a granularity
Flip three coins and ask for the probability of exactly two heads. Outcomes as ordered triples: 8 equally likely, of which HHT, HTH, THH qualify, so 3/8. Outcomes as "number of heads": four possibilities, but not equally likely — and answering 1/4 is the error the fine model prevents.
Interactive
Sample space, event, condition
Exact fractions, every outcome listed
Method
Building a probability model
Outcomes first, and choose them fine enough.
1
List the outcomes
As tuples if several things happen. Fine enough that symmetry makes them equally likely.
2
Check exclusivity and exhaustiveness
Exactly one outcome occurs. If two could occur together, the outcomes are not outcomes.
3
Write the event as a subset
"At least one 6" is a specific set of pairs. Writing it out is what makes the count possible.
4
Count and divide
Combinatorics and Counting supplies the counting. The division is the only new step in this lesson.
Sample spaces
Score 0 / 0Choose the answer you can justify.
Two dice, three events
S is the 36 ordered pairs, all equally likely.
A = "sum is 7" {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} |A| = 6
B = "a double" {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)} |B| = 6
C = "first die 4" {(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)} |C| = 6
P(A) = P(B) = P(C) = 6/36 = 1/6
A ∩ B = ∅ a double sums to an even number, never 7
A ∩ C = {(4,3)} P = 1/36
B ∩ C = {(4,4)} P = 1/36
P(A ∪ C) = 6/36 + 6/36 − 1/36 = 11/36
A and B are mutually exclusive and A and C are not, even though all three have the same probability. Exclusivity is a fact about the sets, not about their sizes — and “Independence” shows that independence is a third, different property again.
Common mistakes
1
Choosing outcomes too coarseSums of dice, numbers of heads and unordered pairs are all legitimate descriptions and none of them is equally likely. Keep the outcomes fine.
2
Assuming equal likelihood without a symmetryA loaded die has the same sample space and different probabilities. The assumption belongs to the model and should be stated.
3
Treating mutually exclusive as independentThey are nearly opposite. Two exclusive events with positive probability are strongly dependent: knowing one occurred tells you the other did not.
Check your understanding
✓
Finish when you write the sample space before computing.
For "draw two cards from a standard deck", write down what an outcome is, how many there are, and whether they are equally likely — before touching any event.
Note: For countably infinite sample spaces — "flip until the first head" — the equally-likely model is unavailable and probabilities are assigned directly. “The Geometric Distribution and Waiting Times” does exactly that for the geometric distribution.